Schedule S is R1(A), W2(A), R2(B), W1(B). What does its precedence graph show?
Correct answer B. Edges T1 -> T2 and T2 -> T1, so S is not conflict serializable
Explanation
R1(A) comes before the conflicting W2(A), which gives the edge T1 -> T2. R2(B) comes before the conflicting W1(B), which gives T2 -> T1. The two edges form a cycle, and a schedule is conflict serializable if and only if its precedence graph is acyclic. So no serial order of T1 and T2 is conflict-equivalent to S.
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